Find $\lim_{x\to \frac12}\dfrac{x\cos(\pi x)}{e^{x}-\sqrt{e}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\pi}{2\sqrt{e}}$ (Choice B) B $-\dfrac{1}{2\sqrt{e}}$ (Choice C) C $\dfrac\pi 2$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=\dfrac12$ into $\dfrac{x\cos(\pi x)}{e^{x}-\sqrt{e}}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to \frac12}\dfrac{x\cos(\pi x)}{e^{x}-\sqrt{e}} \\\\ &=\lim_{x\to \frac12}\dfrac{\dfrac{d}{dx}\left[x\cos(\pi x)\right]}{\dfrac{d}{dx}[e^{x}-\sqrt{e}]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \frac12}\dfrac{\cos(\pi x)-\pi x\sin(\pi x)}{e^{x}} \\\\ &=\dfrac{\cos\left(\dfrac{\pi}2\right)-\dfrac\pi 2\sin\left(\dfrac{\pi}2\right)}{\sqrt{e}} \gray{\text{Substitution}} \\\\ &=-\dfrac{\pi}{2\sqrt{e}} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \frac12}\dfrac{\dfrac{d}{dx}\left[x\cos(\pi x)\right]}{\dfrac{d}{dx}[e^{x}-\sqrt{e}]}$ actually exists. In conclusion, $\lim_{x\to \frac12}\dfrac{x\cos(\pi x)}{e^{x}-\sqrt{e}}=-\dfrac{\pi}{2\sqrt{e}}$.